Transcript Capacitance - Stephen F. Austin State University

Capacitance
The Ability to Store Charge
Common Types
Electrolytic
Tanalum
Silver Mica
Parallel Plate Capacitor
d
+
+
+
+
+
+
―
―
―
―
―
―
Plate Area = A
Dielectric
Capacitance Depends on…
 Separation of the plates (d)
1
C
d
 Plate Area (A) [A in m2]
[d in m]
CA
 Dielectric Constant ()
C 
[ in C2/N-m2]
Capacitance
C
A
d
Units :
 C2  2

m
2 
2 2
2
2
Nm
C
m
C
C





3
m
Nm
Nm J
 farad
 
Capacitance depends on the construction of the
capacitor, not on the circuit it is used in.
Alternate Form for C
from the units
2
C
C
C
Capacitanc e 


J
J /C V
Thus,
C = Q/V
Recall, V=W/Q
Example
 What is the capacitance of a capacitor
if the charge is 0.075 C held at V =
400 V
 C = Q/V =0.075 C/400 V
 = 1.875 X 10-4 f = 187.5 mf
Storage of Charge
 Charges flow from the wire to one plate of
the capacitor. The insulator (dielectric)
prevents charges from flowing to the other
plate. Charge accumulates.
 The accumulated charge attracts opposite
charges on the other plate.
 The second plate charge pushes opposite
charges down the other wire.
+
+
+
+
+
+
―
―
―
―
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Capacitors in DC Circuits
 Capacitor in a dc circuit
 Capacitor will begin charging and current
will flow
 When fully charged (Q= CV), current
stops
 Electric field between plates cancels the
field produced by the battery voltage
Dielectric Constant
 Many texts give  in terms of a
fundamental constant (o) and a number
that depends on the material (K)
 = Ko
o = permittivity of free space
= 8.854 X 10-12 C2/Nm2
K = dielectric constant
Typical Values of K
Material
Air (1 atm)
K
1.00059
Vacuum
Ammonia
Glass
Mica
Paraffin Wax
Porcelain
Rubber
1.00000
22
5-10
3-6
2.1-2.5
6.0-8.0
2.5-3.0
Example
 Two rectangular sheets of copper foil 16 X 20 cm are
separated by a thin layer of paraffin wax 0.2 mm thick.
Calculate the capacitance if the dielectric constant for
the wax is 2.4.
K o A
d
A  ( 16 cm )( 20 cm )  320 cm 2  0.032 m 2
C
d  0.2 mm  2 X 10 4 m
( 8.854 X 10 12 C 2 / Nm2 )( 2.4 )( 0.032 m 2 )
C
2 X 10 4 m
C  3.4 X 10 9 f  3400 pf
Combining Capacitance
 Parallel – plates of the capacitors are
at the same voltage as the poles of
the battery.
Voltage across AA’ = BB’ = E
Parallel Capacitance
Q1 = C1E
Q2 = C2E
Q3 = C3E
Total Charge stored is…
QT = Q1 + Q2 + Q3
QT = E(C1 + C2 + C3) = CTE
CT = C1 + C2 + C3
Series Capacitance
The same current flows through each capacitor
Current is charge/time, so Q1 = Q2 = Q3 = QT
Series Capacitance
 Kirchhoff’s Voltage Law
 E = V1 + V2 + V3
QT Q1 Q2 Q3



CT C1 C2 C3
QT QT QT QT



CT C1 C2 C3
1
1
1
1



CT C1 C2 C3
Try it
 Adding Capacitance
Example
Three capacitors of 2.0, 3.0, and 5.0 mf
are connected in parallel to a 12 V
source.
1. Find the charge on each capacitor.
2. Find the total charge of the
combination.
3. Find the total charge if the same
three are connected in series to the
12 V source.
Solution (Parallel connection)
1. Q1 = VC1 = (12 V)(2.0X 10-6f) =24 mC
Q2 = VC2 = (12 V)(3.0X 10-6f) =36 mC
Q3 = VC3 = (12 V)(5.0X 10-6f) =60 mC
2. QT = Q1 + Q2 + Q3 = (24+36+60)mC
= 120 mC
Solution (part 3) Series connection
1
1
1
1



CT C1 C2 C3
1
1
1



2.0 mf 3.0 mf 5.0 mf
15  10  6 31


30
30
30
CT 
mf
31
QT  VCT  ( 12V )(
QT  11.6 mC
30
X 106 f )
31
RC Circuits
We can apply Kirchhoff’s Laws to this circuit also
E = VC + VR = Q/C + IR
Charge Form
 For a dc voltage flowing for a time t
 I = Q/t
 E = Q/C + QR/t = Q(1/C + R/t)
E
E
Q

1 R
t  RC
C
t
Ct
ECt
Q
t  RC
Current Form
 Q = It
 E = Q/C + IR = It/C + IR = I(t/C + R)
E
I
t  RC
C
EC
I
t  RC
Time dependent forms
 Notice that RC must have units of time –
most books call this t = RC
 Current
 Voltage
E t / t
i e
R
V  E( 1  e
t / t
)
10
I vs. t
9
8
6
5
4
3
2
1
12
0
0
1
2
3
4
V vs. t
10
Time
Current
Current
7
8
6
4
2
0
0
1
2
Time
3
4
Saving Links
 Charging Circuits
 RC Circuit Applet